# Trying to understand the relationship between voltage, current, and power

Please forgive this ultra-noob question, but I'm a bit confused. I've been reading about how neither voltage nor current alone constitute power in a DC circuit. Instead, power is a combination of both voltage and current. Alright... But if current is the flow rate of electrons through a circuit and if it's true that voltage causes current, then doesn't current alone already imply voltage? Why doesn't this mean then that current on its own indicates power?

Muchas gracias.

• I think your thinking is partially correct. If a current is flowing, this automatically implies a potential difference and hence power - even in cases where the current is not generated by the potential difference itself but by other means, such as thermally. Current on its own can't indicate power, though, since if there are moving charges there will always be a potential difference involved. – Bitrex May 23 '11 at 20:47
• Can you provide a quote from wherever you read this from? Just so we can rule out any mis-interpretation. My guess is that whoever made that statement was trying to say that knowing only the voltage or the current alone is not enough to calculate the precise Power. You are correct in that current does imply power but it says nothing about how much power. – Jon L May 23 '11 at 21:00
• "doesn't current alone already imply voltage?" Yes. But how much? Depends on the resistance. The power does also. – endolith May 23 '11 at 21:23
• @Jon- I got most of this from Kuphaldt's 'Lessons in Electric Circuits, Volume I - DC, Ohm's Law', which you may have seen at <allaboutcircuits.com/vol_1/chpt_1/index.html>. – echobase May 23 '11 at 22:59
• @endolith - "doesn't current alone already imply voltage?" No! In a superconductor you can have current without a voltage drop – stevenvh May 26 '11 at 7:09

I'm assuming we are talking in terms of DC here to keep things simple.

Voltage is a measure of how much a single electron's energy changes across the circuit. Current is how many electrons are flowing through a circuit. Multiply those two values to get the rate of energy transfer, or power. If all you have is a measure of current, you know how many electrons are flowing through the circuit, but not how much their energy is changing.

You are correct that voltage induces current, and that voltage and current are interrelated. Ohm's law is the simplest case of that. You can derive a circuit's power from current and it's resistance using known formulas. And even more simply, if you know current is flowing you know the circuit is dissipating power, even if you don't know how much (as the other commenters have said)

• You nailed it.. – 0xakhil May 24 '11 at 18:14
• We increase voltage at the cost of current during power transmission to reduce I^2xR loss. Why doesn't V^2/R equation to calculate power applicable there? – 0xakhil May 24 '11 at 18:29
• It does. But the V in that equation is the V between the two ends of the line (or the difference in V-to-ground at the two ends of the line), not the transmission voltage itself. If you doubled the transmission voltage (halving the current and dividing the power loss by 4), you'd also expect to see the voltage drop along the line get halved, and the v^2/R would produce the same result as I^2*R. – JRobert May 24 '11 at 20:27
• Replace 'electron' with 'electric charge'. It's $VI = \frac{J}{C} \frac{C}{s} = \frac{J}{s} = W$, where J is joules of energy and C is coulombs of charge. If you short circuit a high AC voltage with your body, a large current flows through you, but it's a flow of ions, not electrons. – Eryk Sun May 25 '11 at 2:13

@echobase: though it's an imperfect analogy, it is sometimes helpful to compare voltage to "water pressure", current as "rate of flow" and power as "force of flow".

To flesh that out: imagine you have a water wheel, and you need power to turn it.

If you have really high water pressure, but it's coming through a tiny hole (high voltage, low current), you won't get much power. If you have a big pipe but low pressure (high current but low voltage), you won't get much power either. It takes both pressure (voltage) and flow (current) to get work done (power).

If you're attracted to the water-as-electricity analogy, you'll probably enjoy this link from Georgia State University

Power $P = I^{2} R$. This means that to have power you need both current and some resistance. Now even a copper wire has a resistance, so there will be power dissipation in the wire, but often both resistance and (thus) power are neglectable.
There is a special condition, though, when you have a current in a superconductor. In that case resistance is indeed zero, so there will be no power dissipation, even if there is a nonzero current. Since $V = I R$ the current won't generate a voltage either.

With DC electronics, there are rarely situations where current and voltage don't mean power. (I can't think of any off hand.)

With AC electronics, you can have current flowing due to capacitive or inductive reactance. These can cancel out and not truly consume any power from the source. Any purely resistive load, P = IV should be true. However, when you bring in power factor, things change.

That may be more complex than you are thinking. If you are talking only DC power, then current can imply power. You can take V = IR and P = IV and get power from any two factors. You cannot get power just from current. How much power is consumed if 1 A is flowing through a huge diameter wire? Almost nothing. It is just on its way to power something else. How much power is consumed if 1 A is flowing from a 10 V battery or across a 0.1 Ohm resistor? Now you can say 10 W.

It isn't that you NEED voltage, but that without two of the three items with DC electronics, you can't tell how much work is being done.

I'll try to add a practical example to this:

One situation where you can have current and no voltage in DC electronics is with solar cells. As light hits a solar cell an energy transfer occurs that causes a current, if the cell has a wire attached connecting the positive and negative (ground) terminals then this current will travel along the wire and back into the cell. There can be quite a lot of current but no voltage, as the wire has no resistance.

If we go to the other extreme and remove the wire from the solar cell, the energy transfer from light to electricity still occurs but the current gets "stuck" at the terminals of the device. This current builds up (almost instantly) to a set voltage. In this case the voltage can be measured but there is no current flow because the resistance between the terminals is infinite.

Now, if we attached a resistor across the solar cell (lets make it a big one) some of the current is able to flow through it and back to the cell. Assuming the resistance is large the current flow will be small and the voltage won't reduce very much.

If we added a much smaller resistance then it is really easy for the current to flow out of the solar cell and back to the other side. Not all of the current can get through at once though, so we can still see a small voltage across the solar cell (caused by some "stuck" current).

By changing the resistance we put across the solar cell we can change the relationship between the current and voltage, as you have seen power = current x voltage so if we have 0 current and lots of voltage (or the other way round) you can see how we have 0 power. By adding resistance we get a little bit of both and as such we get power. That power is quite literally heat lost in the resistor in case you were wondering where the energy goes.

NB: This is a hugely simplified example and explanation but if you are as noob as you claim then this should do the trick.

Another water example. voltage is the water in the tank in the water tower, current is the rate of flow of that water through the pipe to the ground, power is the current slamming into resistance (the inertia from the mass could be the voltage and the rate the mass is moving the current). from ohms law as stated in another answer you can derive power from current and resistance alone (p=i*i*r).

think about static electricity and the shock you get from it, the voltage numbers are huge, but the power isnt there, voltage or current by itself is not enough you need a supply of both to really do something and that is why the combination is needed to get power. A really fast drop of water isnt going to do much, you need some voltage behind it. Think of a water balloon and a water bed, they may fall at the same rate but only one will total your car when it hits the street. The combination mass being the voltage and current being the rate the mass is moving, voltage being the multiplication of the two very similar to the way that that force is mass times acceleration, one without the other doesnt do much.